>> Can someone please tell me how to prove that the real vector space of>> all sequences of real numbers has uncountable dimension?>> Suppose the n't term of some sequence is s_n.>> Then the set S of vectors 's' such that s_n = O(1/n^2)> is also vector space ...>>> Hint:> s_n = O(1/n^2) means "Exists K, a real number, |s_n| <= K * (1/n^2) " .>> So S is a real vector space.> All elements of S are elements of the real Hilbert space l^2> of square-summable sequences of reals .>> S is vector subspace of l^2, the real infinite-dimensional Hilbert> space. [ infinite dimensional: with Hilbert spaces, one normally> classifies spaces by the cardinality of a Hilbert basis of> the Hilbert space].>>> By what appears below, a Hamel basis (vector space basis)> of the infinite dimensional> real vector space l^2 does not have a countable basis.> So S does not have a countable basis.>>> Background at PlanetMath> =========================>> There's a Panet Math entry with title:> "Banach spaces of infinite dimension do not have a countable Hamel basis" :> http://planetmath.org/encyclopedia/ABanachSpaceOfInfiniteDimensionDoesntHaveACountableAlgebraicBasis.html>>> Usually, when nothing is said, statements on Banach spaces> apply irrespective of whether it is a real Banach space> or a complex one ...>> The proof at PlanetMath appeals to the Baire Category Theorem,> http://en.wikipedia.org/wiki/Baire_category_theorem>> "(BCT1) Every complete metric space is a Baire space.">> As I recall, when using the form I know of BCT, we always want> the metric space to be complete.>>> Banach spaces are complete normed spaces .>>>> They cite a Monthly article:> 1 H. Elton Lacey, The Hamel Dimension of any Infinite Dimensional> Separable Banach Space is c, Amer. Math. Mon. 80 (1973), 298.>> ( c = cardinality of the continuum ).>> A Hamel basis is a basis in the sense of vector spaces (only finite sums).>> A Hilbert basis for a Hilbert space :> The Wikipedia article on Orthonormal basis says this about> "Hilbert basis":>> "Note that an orthonormal basis in this sense is not generally a Hamel> basis, since infinite linear combinations are required.">> cf.:>> http://en.wikipedia.org/wiki/Orthonormal_basis>>> Baire space: [at Wikipedia] Any countable intersection of dense open> sets, is itself dense.>> N.B.: There's a way of switching things around by looking> at the complements, which are then closed sets.> [ By De Morgan's laws, if I'm not mistaken ].